I know that the standard pre-emphasis function is $y[n] = x[n]-\alpha\cdot x[n-1]$ and $\alpha$ is often set as $0.97$.
However, when I was learning speech processing using Praat, the pre-emphasis factor in it is $6\ \rm dB/oct$ which means the signal emphasise $6\ \rm dB$ per octave increase. That's what confused me.
I've read lots of questions and answers, most of them are explaining how to calculate the value of $\alpha$ or the $\mathcal Z$-transformation. But What I'm confused is the connection between $\alpha$ and $6\ \rm dB/oct$.
I found this equation in the notes of Praat:
The pre-emphasis factor $\alpha$ is computed as: $$ \alpha = \exp(-2\pi F \Delta t) $$ where the $F$ above which the spectral slope will increase by $6\ \rm dB/oct$, $\Delta t$ is the sampling period of the sound
According to above notes, it seems that the $\alpha$ only influences the "beginning frequency" of pre-emphasis but not "how much it changes".
Hope I explained myself clearly enough.